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  • Classical Mechanics Problem 1: Central Potential Solution

    a) Integrals of motion for a central potential V (r):

    Angular Momentum L = rvt = r2φ̇ Energy per unit mass E = 12

    ( ṙ2 + v2t

    ) + V (r) = 12 ṙ

    2 + Veff(r)

    where vt is the tangential velocity and Veff is defined as

    Veff(r) = V (r) + L2

    2r2

    If the orbit is circular, the distance of the test body from the origin is invariant: ṙ = 0, which implies that the body is always at the equilibrium-distance:

    dVeff dr

    = 0 ⇒ dV dr

    = L2

    r3 =

    v2t r

    = rφ̇2

    then

    φ̇ = ωφ = L

    r2 =

    ( 1 r

    dV

    dr

    )1/2

    so for the period we get

    Tφ = 2π ωφ

    = 2π (

    1 r

    dV

    dr

    )−1/2

    b) Write the orbit as in the statement of the problem:

    r(t) = r0 + ²(t) with dVeff dr

    (r0) = 0 and ²2 ¿ r20.

    The energy per unit mass is now E = 12 ²̇ 2 + Veff(r0 + ²), and since ² is small we may

    Taylor-expand the potential as

    Veff(r0 + ²) = Veff(r0) + dVeff dr

    (r0)² ︸ ︷︷ ︸

    =0

    + 1 2

    d2Veff dr2

    (r0)²2 +O(²3)

    so then

    E − Veff(r0) = 12 ²̇ 2 +

    1 2

    d2Veff dr2

    (r0)²2 +O(²3) = const. In the above equation we readily recognize the equation of the simple harmonic oscil- lator with

    ωr = (

    d2Veff dr2

    )1/2

    r=r0

    1

  • and its general solution is

    ²(t) =

    √ E − Veff(r0)

    ωr cos [ωr(t− t0)]

    where t0 is an arbitrary constant.

    Now return to writing ωr in terms of V (r) instead of Veff(r).

    ω2r = d2Veff dr2

    = d2V

    dr2 +

    3L2

    r4 =

    d2V

    dr2 + 3ω2φ =

    d2V

    dr2 +

    3 r

    dV

    dr

    ωr = (

    d2V

    dr2 +

    3 r

    dV

    dr

    )1/2

    r=r0

    = [

    1 r3

    d

    dr

    ( r3

    dV

    dr

    )]1/2

    r=r0

    And the radial period is

    Tr = 2π ωr

    c) Stability is determined by the sign of ω2r . For stability: ω2r > 0, so

    1 r3

    d

    dr

    ( r3

    dV

    dr

    ) > 0

    for the Yukawa-potential

    V (r) = −GM r

    e−kr

    so the condition is

    1 r3

    d

    dr

    ( r3

    dV

    dr

    ) =

    GM

    r3 e−kr

    [ 1 + kr − (kr)2] > 0 ⇒ [1 + kr − (kr)2] > 0

    [ 1 + kr − (kr)2] =

    (√ 5− 1 2

    + kr

    ) (√ 5 + 1 2

    − kr )

    > 0

    which is satisfied only if

    kr <

    (√ 5 + 1 2

    )

    Therefore circular orbits are unstable for

    kr >

    (√ 5 + 1 2

    )

    ¥

    2

  • d) The outermost stable circular orbit is at

    r0 =

    (√ 5 + 1 2k

    )

    its energy per unit mass is

    E = V (r0) + 1 2 (r0ωφ)2 = V (r0) +

    1 2

    ( r dV

    dr

    )

    r=r0

    1 2

    GM

    r0 e−kr0(kr0 − 1) = GM

    r0 e−kr0

    (√ 5− 1 4

    ) > 0

    If r0 is decreased only slightly, E > 0 still and the orbit is absolutely stable ¥ The effective potential for the Yukawa-potential has the form shown in Figure 1.

    r

    Veff

    Figure 1: Effective potential against distance from the origin

    3

  • Classical Mechanics Problem 2: Planar Double Pendulum Solution

    l

    l

    q 1

    q 2

    a) L = T − V The moment of inertia for a uniform rod of length l and mass m is

    I = 1 3 ml2 about one of the ends

    and Ic =

    1 12

    ml2 about the rod’s center

    The kinetic energy term we can decompose into three parts:

    T = T1 + T2,rot + T2,trans

    where T1 is the kinetic energy of the first rod, T2,trans is the translational energy of the center of mass of the second rod and T2,rot is its rotational energy about its center of mass. Then

    T1 = 1 6 ml2θ̇21

    T2,rot = 1 24

    ml2θ̇22

    and T2,trans =

    1 2 m

    ( ẋ2c + ẏ

    2 c

    )

    where xc and yc are the coordinates of the second rod’s center of mass, so

    xc = l sin θ1 + l

    2 sin θ2

    yc = −l cos θ1 − l2 cos θ2

    from which

    ẋ2c + ẏ 2 c = l

    2

    [ θ̇21 +

    1 4 θ̇22 + θ̇1θ̇2 (sin θ1 sin θ2 + cos θ1 cos θ2)

    ]

    4

  • The potential energies are simply Vi = mgyc,i, where yc,i are the vertical coordinates of the rods’ centers of mass. Since both rods are uniform, yc,i are simply the coordinates of the centers. Thus,

    V1 = −mg l2 cos θ1; V2 = −mg (

    l cos θ1 + l

    2 cos θ2

    )

    The full Lagrangian is then

    L = T1 + T2,rot + T2,trans − V1 − V2

    = ml2 [ 2 3 θ̇21 +

    1 6 θ̇22 + θ̇1θ̇2 cos (θ1 − θ2)

    ] + mgl

    [ 3 2

    cos θ1 + 1 2

    cos θ2

    ]

    b) Expand the Langrangian from part a) for small angles. The only function we have to deal with is

    cos θ = 1− 1 2 θ2 +O(θ4)

    Since we are going to look for normal modes with θj = θ̂j exp (iωt), where the θ̂j ¿ 1, we immediately see that in the term θ̇1θ̇2 cos (θ1 − θ2), the θ-dependence in the cosine can be dropped, because even the first θ-dependent term gives a fourth order correction. Then the approximate Lagrangian is

    L = ml2 [ 2 3 θ̇21 +

    1 6 θ̇22 + θ̇1θ̇2

    ] −mgl

    [ 3 4 θ21 +

    1 4 θ22

    ] + const.

    The Euler-Lagrange equations are

    d

    dt

    ∂L

    ∂θ̇j =

    ∂L

    ∂θj

    so in the specific case:

    4 3 θ̈1 +

    1 2 θ̈2 +

    g

    l

    ( 3 2 θ1

    ) = 0

    1 2 θ̈1 +

    1 3 θ̈2 +

    g

    l

    ( 1 2 θ2

    ) = 0

    if we now look for normal modes, as mentioned, the above set of equations takes the form [ (

    4 3ω

    2 − 32 gl )

    1 2ω

    2

    1 2ω

    2 (

    1 3ω

    2 − 12 gl )

    ] [ θ̂1

    θ̂2

    ] = 0

    Non-trivial solutions exist if the determinant of the matrix on the left is zero. Denoting ω2 = λg/l, we can write this condition as

    ( 4 3 λ− 3

    2

    ) ( 1 3 λ− 1

    2

    ) − λ

    2

    4 = 0,

    5

  • that is 7 36

    λ2 − 7 6 λ +

    3 4

    = 0

    whose solutions are λ± = 3± 6√

    7 ,

    so finally

    ω± = [(

    3± 6√ 7

    ) g

    l

    ]1/2

    c) To sketch the eigenmodes, find eigenvectors of the matrix in part b).

    • ω2 = λ−g/l (low-frequency mode)

    θ̂2 = (

    3 λ − 8

    3

    ) θ̂1 =

    1 3

    ( 2 √

    7− 1 )

    θ̂1

    (2 √

    7− 1)/3 > 0 and real, therefore the two pendula are in phase; • ω2 = λ+g/l (high-frequency mode)

    θ̂2 = (

    3 λ − 8

    3

    ) θ̂1 =

    1 3

    ( −2 √

    7− 1 )

    θ̂1

    (−2√7− 1)/3 < 0 and real, therefore the two pendula are perfectly out of phase.

    a b

    Figure 2: The low- (a) and high-frequency (b) normal modes of the planar double pendulum.

    6

  • Electromagnetism Problem 1 Solution

    a) Normal modes are products of harmonic standing waves in the x, y and z directions. For their frequencies, we have

    ω = c √

    k2x + k2y + k2z = c [(πnx

    a

    )2 +

    (πny b

    )2 +

    (πnz b

    )2]1/2 ; nx, ny, nz ∈ Z+

    Since a > b, the lowest frequency has nx = 1 and either ny = 1, nz = 0 or ny = 0, nz = 1 (note that ny = 0, nz = 0 does not satisfy the boundary condition E‖,at wall = 0). Since we are told to pick the mode with ~E ‖ ŷ, the boundary conditions require

    ~E(r, t) = E0ŷ sin πx

    a sin

    πz

    b cosωt

    The magnetic induction we can get from Faraday’s Law:

    ∂ ~B

    ∂t = −c

    ( ∇× ~E

    ) = x̂c

    ∂Ey ∂z

    − ẑc∂Ey ∂x

    = ( x̂cE0

    π

    b sin

    πx

    a cos

    πz

    b − ẑcE0 π

    a cos

    πx

    a sin

    πz

    b

    ) cosωt

    ~B(r, t) = −πcE0 ω

    ( x̂

    b sin

    πx

    a cos

    πz

    b − ẑ

    a cos

    πx

    a sin

    πz

    b

    ) sin ωt

    where the frequency ω is (by the argument above)

    ω = πc (

    1 a2

    + 1 b2

    )1/2

    b) At a boundary of media, the discontinuity in the normal component of the electric field is 4π times the surface charge density σ, so

    Ey(x, 0, z) = 4πσ

    σ(x, 0, z) = E0 4π

    sin πx

    a sin

    πz

    b cosωt

    σ(x, b, z) = −σ(x, 0, z) and

    σ(0, y, z) = σ(a, y, z) = σ(x, y, 0) = σ(x, y, b) ≡ 0 Similarly, at the boundary of media the discontinuity of the tangential component of the magnetic field is given by the surface current ~κ

    n̂× ~B = 4π c

    7

  • where n̂ is a unit vector normal to the surface, so

    ~κ(x, 0, z) = c2E0 4ω

    ( ẑ

    b sin

    πx

    a cos

    πz

    b +

    a cos

    πx

    a sin

    πz

    b

    ) sin ωt